The electric field of an infinite plane above its surface is $E=\frac{\rho}{2\epsilon_0}$, where $\rho$ is the surface charge density and $\epsilon$ is the permittivity of free space. If we have two planes, I know that we can find the total electric field by superposition.
My question is, given two planes of two different charge density, say $\rho$ for the bottom one and $-2\rho$ for the top one. Obviously, the field in between the planes is perpendicular to them.
But using Gauss law, if we enclose only one of the surfaces, we get that the electric field between the planes should only depend on the charge density of that plane, but by superposition we have that $E=\frac{3\rho}{2\epsilon_0}$, which is not equal to any one field generated by one plane.
What went wrong in my application of Gauss law?
Best Answer
In the original application of Gauss's law to one infinite plate of charge, you have 2 Gaussian surfaces over which you have to integrate - the two faces of the pill box parallel to the plate (the 4 faces perpendicular to the plate you don't have to integrate over because the E-field will be parallel to those surfaces and so the flux will be 0). You can assume that these 2 faces have the same electric fields moving across them (you set up the Gaussian surface so that these 2 faces are equidistant from the plate) and so you get the standard result for electric field due to an infinite plate.
When you have 2 infinite plates, you can no longer say that the 2 faces have the same electric fields moving across them since you can't have the two faces being equidistant from both plates at the same time. So, Gauss's law is still valid, but you just can't factor out the electric field in the equation like you could with one plate. The easy way to solve the problem then is by using superposition.
Gauss's law is always correct, you just need a high degree of symmetry to easily apply it. If the symmetry is broken, the law becomes harder to apply.
EDIT: As per request: Gauss's law says that the total integral over a closed surface of the electric flux is equal to the enclosed charge. That's all it says. For charges which are not enclosed by the Gaussian surface, they don't affect this total integral because any electric flux going into the gaussian surface will also exit out of that Gaussian surface. Gauss's law doesn't say that external charges don't affect the electric field itself on that surface.